Question

A piece of aluminum metal at an initial temperature of 97.3°C was placed in a calorimeter containing 49.0g of water at an initial temperature of 22.0°C. The two were allowed to come to thermal equilibrium and the final temperature was 24.1°C. The specific heats of water and aluminum are 4.184 J/g °C and 0.902 J/g °C, respectively. What was the mass of the Al piece that was added?

Answer 1

Answer : The mass of the Al piece that added was, 6.52 grams.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of Al =
`0.902J/g^oC`

`c_2` = specific heat of water =
`4.184J/g^oC`

`m_1` = mass of Al = ?

`m_2` = mass of water = 49.0 g

`T_f` = final temperature of mixture =
`24.1^oC`

`T_1` = initial temperature of Al =
`97.3^oC`

`T_2` = initial temperature of water =
`22.0^oC`

Now put all the given values in the above formula, we get

`m_1* 0.902J/g^oC* (24.1-97.3)^oC=-49.0g* 4.184J/g^oC* (24.1-22.0)^oC`

`m_1=6.52g`

Therefore, the mass of the Al piece that added was, 6.52 grams.